How do you write the partial fraction decomposition of the rational expression # 1/(x^3-6x^2+9x) #?

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Answer 1 · 2015-12-14T03:53:53

#1/(9x)-1/(9(x-3))+1/(3(x-3)^2#

Factor the denominator.

#1/(x(x-3)^2)=A/x+B/(x-3)+C/(x-3)^2#

#1=A(x-3)^2+B(x^2-3x)+C(x)#

#1=Ax^2-6Ax+9A+Bx^2-3Bx+Cx#

#1=x^2(A+B)+x(-6A-3B+C)+1(9A)#

Thus, #{(A+B=0),(-6A-3B+C=0),(9A=1):}#

Solve to see that #{(A=1/9),(B=-1/9),(C=1/3):}#

Therefore,

#1/(x^3-6x^2+9x)=1/(9x)-1/(9(x-3))+1/(3(x-3)^2#